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Why Numerical Method?

CHAPTER 6

THERMODYNAMIC

PROPERTY RELATION

02/05/2012 Lecture on Thermodynamic Property Relation by Tariku Negash (BED)

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6.1 Introduction

In the solution of engineering problems it is essential to be able to determine the

values of the thermodynamic properties.

While some can be directly measured such as P, v, T others such as u, h, s are

not directly measured

so, one of the major tasks of thermodynamics is to provide basic equations to

evaluate the above un measured properties from measurable property data.

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6.2 Fundamentals of Partial Derivative

Mathematical test is available from partial differential calculus todetermine whether the total differential of the function is an exact

differential.

x=x(y, z);

6.16.2

Where: -

6.3

dz

z

xdy

y

xdx

yz

NdzMdydx

andy

xM

z

yz

xN

yxx

xNand

yxx

zM

zy

22

;

zyy

N

z

M

3


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6.3 The Maxwell Relation

The Maxwell relation is equations that relate the partial derivative ofP,v ,T and S of a simple compressible substance to other.

They are obtained from the four Gibbs equations by exploiting the

exactness of the differential of the thermodynamic properties. Two of them are:-

6.4

6.5

6.31 The Gibbs and Helmholtz Relation

6.6

6.7

Tsua Tshg

PdvTdsdu vPdpTdsdh

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Cont

Helmholtz function

6.8

Derivativeboth sides of Eq. (6.6 ),then6.9

Substitute Eq. (6.4), then

6.10

Gibbs Function

Derivativeboth sides of Eq. (6.7),then6.11


sdTTdsduda

sdTPdvda

sdTTdsdhdg

sdTPdvda

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Cont...

Derivativeboth sides of Eq. (6.7), then6.12

Substitute Eq. (6.5) in terms of dh, then

6.13

For summarize

6.14

6.15

6.16

sdTTdsdhdg

sdTTdsdhdg

PdvTdsdu

vPdpTdsdh

sdTPdvda

sdTTdsdhdg

6


7/4002/05/2012 Lecture on Thermodynamic Property Relation by Tariku Negash (BED)

Cont...

The quantities u,h, and g are properties and therefore they

should have exact differentials. Applying Eq. ()

6.17

6.18

6.19

6.20

This set of equation is referred to as theMaxwell relations.

sv v

T

s

P

sv P

T

s

v

Tv vs

TP

TPP

s

T

v

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6.4 Generalized relation for

The state postulate established that the state of a simple

compressible system is completely specified by two

independent, intensive properties.

6.41 Internal Energy Changes (du)

We choose the internal energy to a function of T and V, that isand take its total differential.

u=u(T, v) and take its total differential.

6.21

Or ,6.22

pvandccdsdhdu ,,,,

dvv

udT

T

udu

Tv

dvv

udTcdu

T

v

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Cont


Now let us express s, as s = (T,v). Therefore,

6.24

Using Eq. (6.24), the relation, can be written as

6.25

Equating the coefficient of dT and dv in Eq. (6.24) and (6.25)

and then second relation above, we obtain

6.26

Using the Maxwell relation of Eq.(6.17) and the second relation

above, we get

dvvsdT

Tsds

Tv

dvPv

sTdT

T

sTdu

Tv

T

c

T

s v

v

P

v

sT

v

u

TT

9


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Cont


6.27 With Eq. (6.22), yields and the change in u of a simple compressible

system associated with a change of state from (T1,v1) to (T2,v2) is

given by

6.28

6.42. Enthalpy Changes (dh)

Let h=h(T,P). Therefore,

,or 629

6.30

P

T

PT

v

u

vT

dvPTPTdtcuu

T

v

v

T

Tv

2

1

2

112

dPv

hdT

T

hdh

TP

dPP

hdTcdh

T

P

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Cont...

Assuming s=s(T,P) we have

6.31

Using this into the relation dh=Tds + vdP results in

6.32

From Eq. (6.30) and (6.32), we get

and, 6.33

Using the Maxwell relation of Eq. (6.19) and the second relation

above, we get

6.34

dPP

sdTT

sds

TP

dPv

P

sTdT

T

sTdh

TP

T

c

T

s P

P

v

P

sT

P

h

TT

dvT

PTvdTcdh

P

P

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Cont..

and, 6.37

By Maxwells relation of Eq. (6.19), we have

6.38

Therefore,

6.39

6.44. Specific Heats Cv and CP

6.40

Check it by your self ?


dPP

s

dTT

s

dsTP

Tc

T

s P

P

PTT

v

P

s

dPT

vdT

T

cds

P

P

TP

vPT

P

T

vTCC

2

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6.5 The Joule-Thomson Coefficient

When a fluid passes through a restriction such as porous plug,

a capillary tube or an ordinary valve, its pressure decrease, but

the enthalpy of the fluid remains approximately constant

during such a throttling process.

Fig (6.1). The temperature of a fluid may increase, or decrease, or

remain constant during a a throttling process

Joule-Thomson Coefficient, defined as (h=constant)

6.41


hP

T

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Cont

Thus the Joule-Thomson Coefficient is measure of a change in

temperature with pressure during a constant-enthalpy process.

Notice that if

temperature increase

temperature remain constant

temperature decreaseOn the T-P diagram can easily constructed from temperature and

pressure measurements along throttling processe.


0

0

0

JT

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Cont

Some constant enthalpy lies on the T-P diagram pass through a

point of zero slope or zero Joule-Thomson coefficient.

Inversion line:- is a line that passes through this point (zero slope)

Inversion temperature: - is a temperature at point where a

constant enthalpy line intersects the inversion line.

Maximum Inversion Temperature:- ism the temperature at theintersection of the P=0 line (ordinate) and the upper part of the

inversion line.

6.5.1 A general relation for Joule-Thomson Coefficient

From the previous Maxwell relation equation, that is6.42

Partial derivative of each term with respect to P and h=constant

6.43


dPT

vTvdTcdu

P

P

16

PPh

JTT

vTv

cP

T 1


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Cont

Table (6.1) Joule-Thomason Coefficient for air, in K/bar

Fig. (1). Joule-Thomson temperature, k


Temperature, k Pressure, bar1 50 100 150 200 500

150 0.804 0.644 0.132 0.042 0.010 -0.037200 0-500 0.412 0.29 0.168 0.092 -0.026250 0-336 0.273 0.215 0.155 0.105 -0.016275 0-280 0.227 0.159 0.136 0.097 -0.014300 0-234 0.19 0.112 0.117 0.085 -0.013350 0.165 0.133 0.107 0.083 0.061 -0.015400 0.116 0.091 0.072 0.055 0.040 -0.020500 0.051 0.037 0.025 0.016 0.007 -0.031800 -0.030 -0.034 -0.037 -0.040 -0.043 -0.055

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Cont

By setting Eq. (6.49) equal to zero, we find that the general

criteria for the inversion curve in terms of PvT data is

, or 6.44

A semi quantitative explanation of the trends shown in Fig. (1 ) is

found by exploding the equation of state Pv= ZRT inEq.(6.49). The result is

6.45

Thus cooling occurs when the sign is positive on the basis of the

Z chart

Fig( 2 ) The inversion curve on reduced coordinate


PCP

JTT

z

P

RT

2

18

T

v

T

v

P

r

r

Pr

r

T

v

T

v

r


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6.6 Residual Property Function

Even after the development of a P, v, T state based on

experimental data several other data must be considered.

The specific format emphasis to evaluate u, h, s and other

thermodynamic property data depend on which the P, v, T

equation is explicitly in specific volume or pressure.

In practice equation of state are developed in either the format

of v=f(P,T) or P=f(T,v) or P=f(T,)

Lets any specific property (y) are residual function yR is

defined as

yR = y*-y or yR = y-y*

Where:- y= the desired value of y at (T, P) andy*= the function would have at the same temperature and pressure value

if the fluid in an ideal gas.

(*)= is hypothetical one, but the introduction of it proves is to be quite useful.


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cont

6.61 Evaluation of Enthalpy Changes

Consider a change of state for a real gas as represented by state1 and 2 on the p-h and h-s diagram in Fig. (3). In terms of

these Eq. (6.53) may be written as

6.46

This is given by the integral of Cp,odT. Hence the above equationreduces to

6.47

6.62 Evaluation of Entropy Changes

A general equation for the entropy changes between two realgas states has the same format as Eq. (6.36), namely,

6.48


11222211

,*

,*

,2

*

2,1

*

112 TPTPPTPThhhhhhhh

dTchhhhT

TP

RR

2

10,2112

1,12,22211

**

,2

*

2,1

*

112 TPTP ssssssss PTPT

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Cont

This given by the well-known ideal-gas equation.

6.49

P-h and h-s diagrams illustrating the use of the residual function

in evaluating property changes between two real-ideal gases.

When we modify Eq. (6.36) and Eq. (6.38) in the followingmanner one may select state 1 in this equations to be ideal gas

reference state with an enthalpy ho* and an entropy so

* at Toand Po. so, h1

R and s1R are zero.

Therefore Eq. (6.36) and Eq. (6.38) becomes in general6.50

6.51


1

2,

2112 ln2

1 PPRdT

Tcssss

T

T

oPRR

andhdTchh RT

ToPo

o

,*

R

o

T

T

oP

o sP

PRdT

T

css

o

ln

,*

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6.7 Residual Properties and the Gibbs Function

From the first and second law consideration the property most closely

related to temperature and pressure in the Gibbs unction (recall that

Rather than develop an expression for dgR as a function of dP and dT

better to work with its dimensionless format, d(gR/RT). Recall the

mathematical identity.6.52

Consequently, , R is constant number 6.53

But, at constant composition, and g=h-Ts substitution yields

6.54


).sdTvdPdg

222 x

ydx

x

dy

x

ydx

x

xdy

x

yd

2RT

gdT

RT

dg

RT

gd

dTRT

hdp

R

v

RT

gd

2

22


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Cont

From yR=y*-y=yig-y. thus subtraction of Eq. (6.54) yields

6.55This result is a fundamental residual property relation for system

of constant composition.

The final step to express gR/RT as function of P, v, and T.

(Constant T) 6.56

Where:-

VR = v*-v = 6.57

For a constant pressure Eq. (6.55)

(Constant P) 6.58


dTRT

h

dpR

v

RT

g

d

RRR

2

dPRT

v

Pdp

RT

v

RT

g RPRR

00

1

vP

RT

P

RR

T

RTgT

RT

h

/

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Cont

From the definition of g we may write gR = hR-TsR

Therefore, in dimensionless format the residual entropybecomes

6.59

6.60


P

RR

T

RTgT

RT

h

/

RTg

T

RTgT

RT

g

RT

h

R

s RRRRR

/

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6.8 Residual Properties and the Helmholtz Function

As most equations are the form P=P(v,T), the Helmholtz

function, a=a(v,T) also looks to be convenient to derive

residual property relations.

da = -Pdv-sdT or da = -Pdv at T = Contant

6.61

To eliminate the difficulty of infinite limits on the lower andupper bound add and subtract the integral of RT/v as follows:

6.62

Since Z was also defined as Z=vact/videal then Z=v/v* and this will

give

6.63


v

vv

v

R Pdvdvv

RTPdvaaa

**

*

vv

vv

v

v

v

R

v

vRTdv

v

RTPdv

v

RTdv

v

RTP

dvv

RTdv

v

RTdv

v

RTPaaa

*

*

ln*

*

CTZRTdvv

RTPaaa

vR

ln

*

25


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Cont

To make it dimensionless divide by RuT

6.64

To get the residual functions for s and h, start with

da==Pdv-sdT and noting that s=(a/T)v, hence

6.65Substitution of (a*-a) and taking the derivative with respect to T

at constant v finally gives

6.66

For enthalpy residual function, start with h = u + Pv = a+Ts +Pv

since a=u-Ts Then hR = h*- h = (a*- a) + T(s*-s) + P*v*- Pv


CTZdvvTRP

TRTRaa

TRa v u

uuu

R

ln1

*

vaaTss )(**

ZRdv

v

R

T

P

RR

ss v u

vuu

ln1

*

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Cont

Where P*v*=RT

Substitutions for (a

*

-a) and (s*

-s) will finally give6.67

Using g=a+Pv and u=h-Pv the residual functions for g and u

can be determined from gR

/RuT=aR

/RuT+1-Z anduR/RuT=h

R/RuT + Z-1 All the above equations require

P=P(v,T)


11

*

ZdvPT

PT

TRTR

hh v

vuu

27


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6.10 Properties of the Saturation State

It deals with liquid-vapor equilibrium of pure substances.

Fundamental relationships among and approximate evaluationtechniques for the basic properties P, T, u, h, s, a, and g are

sought.

During phase change the heat supplied is called latent heat of

evaporation, hfg

and the change in entropy is given by hfg

/Tsat

.

The combination gives

hfg-Tsfg=0 or h-Ts=0 (Phase change)

From the definition of Gibbs function, g=h-Ts it follows

gT = h - Ts = 0 for a phase change org =g or for the liquid-vapor phase gf=gg

The equality of g for each phase is the criterion for phase

equilibrium.

02/05/2012 Lecture on Thermodynamic Property Relation by Tariku Negash (BED)28


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Cont

From the relationship dg=vdP-sdT

6.68

Since v and s change discontinuously, the derivatives also change

discontinuously. cP=T(s/T)P in the mixture region is infinite

while it has finite values at single phase points.For changes of dT and dP on the two phase equilibrium system at

the initial state (i) giL=gi

V.

At the final equilibrium position

giL

+dgL

=giV

+dgV 6.69

Thus for the change dT dgL = dgV


PT T

gsandP

gv

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Cont


Fig(4) Process involving a differential change in saturation temprature and pressure

Using the expression for dg will give

vL dP sL dT = vV dP sV dT 6.70

And upon rearranging

6.71fg

fg

LV

LV

sat v

s

vv

ss

dT

dP

30


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Cont

Since h = Ts for a phase change

6.72

At low pressures ideal gas behavior can be considered and also

vf


31/40

Cont

This can be seen by inserting v=(ZgZf)RT/P = Z (RT/P) in

the Clapeyron equation and gives the modified Clapeyron

equation as

6.75

The ratio hfg/Zfg tends to remain constant with a minimum

value at Tr

= 0.85. Using Watsons hfg

=(1-Tr

)0.38 and Lileys

Zfg= (1-Tr)0.38 Results in

6.76

Fig ( 5) process involving a differential change in saturation temperature and pressure


Td

RZ

hPd

fg

fgsat 1ln

Td

T

dT

T

dT

RPd

sat 1ln22

32

6 10 1 C i


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6.10.1 Vapor-Pressure Correlations

According to Clausius-Clapeyron equation vapor pressure can be

fitted as

Modification is made on the above equation by using the normal

boiling point Tb (at 1bar or 1 atm) and Tc and Pc to determine

the constants A and B

This will give B=-ATb and A=Tc ln Pc/(Tc-Tb) and the final

equation becomes

To improve the accuracy over a wide range of temperatures, thechange in hfg must be incorporated. A Typical empirical vapor

pressure correlations in use is


RhBtconsAT

BAP fg

sat /tanln

),,,(1ln

ln KTbarsPT

T

TT

PTP b

bc

ccsat

TEDTCTT

BAPsat lnln 2

33

C


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Cont

If a two parameter corresponding state is used on the Clausius-

Clapeyron equation (ln Psat = A-B/T)

It can easily be converted to

6.79

This gives a single straight line approximating the slope for all

substances which is approximately true for simple fluids(


34/40

Cont


Fig (6) application of the law of corresponding state to liquid-vpor equilibrium

for simple fluid

35

C


35/40

Cont

The values for the right hand expressions are given in A-30

Other proposals of the form ln Prsat

=f

(0)

+f

(1)

due to Lee-Kessler

6.81

=/

where:-

=-ln Pc-5.92714+6.09648-1+1.28862 ln -0.1693476

=15.2518-15.6875-1

-13.4721 ln +0.435776

=Tb/Tc , Pc (atm)

Satisfies definition of; curve passes through the critical state; curvature of

the saturation line is zero at the critical state.


6

6

43577.0ln4721.136875.15

2518.15

169347.0ln28862.109648.6

92714.5ln

rr

r

rr

r

sat

r

TTT

TTT

P

36


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Dong and Lienhard proposal:

6.82

Its major advantage over the Lee-Kessler equation is that it

predicts better over a wider range of. Satisfies the definition

of, as well as the critical state condition.


318177.1149408.7(

1137270.5ln r

r

sat

r TT

P

)ln92998.1768769.36

rr TT

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THANK YOU!!

02/05/2012 Lecture on Thermodynamic Property Relation by Tariku Negash (BED) 38

P h d h di


38/40

P-h and h-s diagram


Th i i d d t t


39/40

The inversion curve on reduced temprature


Fi (1) J l Th t t K


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Fig (1) Joule-Thomson,temprature, K

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